Math 127 Homework. Mary Radcliffe. Due 29 March Complete the following problems. Fully justify each response.

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1 Math 17 Homework Mary Radcliffe Due 9 March 018 Complete the following problems Fully justify each response NOTE: due to the Spring Break, this homework set is a bit longer than is typical You only need to turn in those problems marked with (*) 1 In lecture we defined a finite set X to be of size n if there exists a bijection f : [n] X Prove that if X, Y are finite, and there exists a bijection f : X Y, then X = Y Suppose X and Y are finite, so that there exists bijections g : X [n] and h : Y [m] for some n, m N Now, let f : X Y Note that we can construct a function w : [n] [m] by w( = h f g 1 ( Moreover, since h and g are bijections, such a function w is bijective if and only if f is bijective Hence, if there exists a bijection f : X Y, then there exists a bijection w : [n] [m], and hence n = m On the other hand, suppose that X = Y, so n = m Then we can construct a bijection f : X Y by f(x) = h 1 g(x) (*) Let X be a finite set, and suppose there is a surjection f : X Y Prove that X Y 3 (*) Let and Let f : X Y be surjective, and X finite Define a function g : Y X so that for each y Y, g(y) f 1 (y); note that this may not be uniquely defined, but since f is surjective, there always exists such a choice of g(y) for all y Moreover, as f is well-defined, it must be the case that g(y) g(z) whenever y z Therefore, g is injective, and as proved in lecture, this implies that Y X Determine X X 3 X 4 X = {n 1 n 00, n = k k Z}, X 3 = {n 1 n 00, n = k 3 k Z}, X 4 = {n 1 n 00, n = k 4 k Z} We have X = {1, 4, 9, 16, 5, 36, 49, 64, 81, 100, 11, 144, 169, 196} 1

2 and Therefore, we obtain X 3 = {1, 8, 7, 64, 15} X 4 = {1, 16, 81} X X 3 X 4 = {1, 4, 8, 9, 16, 5, 7, 36, 49, 64, 81, 100, 11, 15, 144, 169, 196}, and hence X X 3 X 4 = 17 4 Prove Theorem 41: De Morgan s Law for sets (finite version) The theorem we wish to prove is stated as follows: Let n N For each i [n] let X i be a set, and let Z be a set Then (a) Z\ ( n i=1 X i) = n i=1 (Z\X i) (b) Z\ ( n i=1 X i) = n i=1 (Z\X i) Proof (a) Suppose that z Z\ ( n i=1 X i) Then z Z, and z / n i=1 X i Note that this implies that for all X i, we must have z / X i, as otherwise z would be in the union Therefore, z Z\X i for all i, and hence z n i=1 (Z\X i) Hence, Z\ ( n i=1 X i) n i=1 (Z\X i) For the other containment, suppose that z n i=1 (Z\X i) Then by definition, it must be the case that z Z\X i for all i, so z Z and z / X i for all i Therefore, as z is not a member of any X i, it is also true that z / n i=1 X i Therefore, as z Z, we have z Z\ n i=1 X i Hence, Z\ ( n i=1 X i) n i=1 (Z\X i) By double containment, Z\ ( n i=1 X i) = n i=1 (Z\X i) (b) Suppose that z \ ( n i=1 X i) Then z Z, and z / n i=1 X i Note that if z were a member of X i for all i, then we would necessarily have that z would be a member of the intersection; thus there must exist some choice of i such that z / X i For this choice of i, then, we have z Z\X i Therefore, by definition z n i=1 (Z\X i), and thus Z\ ( n i=1 X i) n i=1 (Z\X i) For the opposite containment, suppose that z n i=1 (Z\X i) Then there exists some i such that z Z\X i, and hence z Z, and z / X i for this specific i Therefore, z / n i=1 X i, since in order to belong to the intersection, z would need to belong to each set But then z Z\ n i=1 X i, and hence Z\ ( n i=1 X i) n i=1 (Z\X i) By double containment, then, Z\ ( n i=1 X i) = n i=1 (Z\X i) 5 You decide to go to Chipotle for a burrito Chipotle has many options: choices of rice, choices of beans, 5 choices of meat, 4 choices of salsa, and 5 other toppings (including queso and guacamole) How many different burritos could you order having exactly 1 rice, 1 bean, 1 meat, salsas, and other toppings? 5 (4 ) ( 5 ) 6 (*) Let X = {(a 1, a,, a n ) a i {0, 1} i} = {0, 1} n These are sometimes called bitstrings of length n

3 (a) Show that there is a bijection between X and {f : [n] {0, 1}}, the set of functions from [n] to {0, 1} (b) Show that there is a bijection between X and P([n]) (c) Determine X (a) Let A = (a 1, a,, a n ) X Define a function F : X {f : [n] {0, 1}} by F (A)( = a k We show that F is injective and surjective To avoid confusing the notation, we will write F A in place of F (A) injective: Let A, B X, with A B Then there must exist some k for which a k b k But then F A ( = a k b k = F B (, and hence F A and F B are not the same function Thus, F is injective surjective: Let f : [n] {0, 1} be a function Define a sequence A by A = (f(1), f(),, f(n)) Clearly F A = f, and hence F is surjective As F is both injective and surjective, it is a bijection (b) Let G : X P([n]) be given by G(A) = {k [n] a k = 1} Note that if A, B X, with A B, then there exists some k such that a k b k ; wlog suppose a k = 1 and b k = 0 Then k G(A) and k / G(B), so G(A) G(B) Hence G is injective For surjectivity, let S [n] Define a sequence A by a k = 1 if k S and a k = 0 elsewise Then G(A) = S by definition, and hence G is surjective Therefore, G is a bijection (c) As we have seen before, P([n]) = n As G : X P([n]) is a bijection, we have X = P([n]) = n 7 (*) Let X and Y be finite sets Define X Y = {f : Y X}, the set of functions from Y to X Prove that X Y = X Y Let S = X X X, where there are Y copies of X As Y is finite, there exists some n so that we can write Y = {y 1, y,, y n } Define a function F : X Y S by F (f) = (f(y 1 ), f(y ),, f(y n )) For the same reason as in the previous problem, F is a bijection, and hence X Y = S Moreover, S is a finite Cartesian product of finite sets, so S = X X X, where there are Y copies of X Hence X Y = X Y 8 (*) Let n, k N with n k Prove, by counting in ways, that k ( n = (n k + 1) ( n Suppose we have n people We form a committee of size k from these people, and then select a leader from that committee Note that there are ( n ways to select the k people, and then k choices of leader, and hence there are k ( n ways to form this committee On the other hand, we could first select k 1 people to be nonleaders, in ( n ways, and then from the remaining n (k 1) = n k + 1 people, choose the leader, which can be done in n k + 1 ways Hence, there are (n k + 1) ( n ways to form this committee Therefore, k ( n k ) = (n k + 1) ( n 3

4 9 Let n, m N with m n Prove that n ( )( ) ( ) n k n = n m k m m k=m Suppose we have a collection of n people We wish to choose m of these people to be Kings, and from the remaining people, we will choose any number to be Queens Everybody else is a Joker We can have any number of Queens, from 0 to n m We can first do this by selecting the m Kings, in ( n ways Then, the remaining people will each be assigned Queen or Joker There are choices for each of these people, so there are n m ways to assign Queens and Jokers Hence, this can be done in n m( n ways On the other hand, we could select a set of k m people from the n people to be Queens or Kings Then we assign m of these k people to be Kings, and the remaining will be Queens For each k m, this can be done in ( n k ( ways, and hence in total there are n ( n k k=m ( such assignments Therefore, n ( n k ) k=m ( ( m = n m n 10 (*) How many subsets of [0] contain a multiple of 4? Prove that your answer is correct In order to contain a multiple of 4, we wish to count subsets that contain either 4, 8, 1, 16, or 0 Given S [0], define X S to be the subsets of [0] that contain S as a subset Notice that X S = 0 S, as we can choose for each element of [0] that is not in S whether to put it in each subset in X S Hence, we wish to count X {4,8,1,16,0} Further, note that X S1 X S = X S1 S, as any element of both will contain all elements in S 1 and all elements in S Hence we have X {4,8,1,16,0} = X 4 X 8 X 1 X 16 X 0 By Inclusion/Exclusion, we obtain X {4,8,1,16,0} = = 5 ( 1) i+1 i=1 i=1 J {4,8,1,16,0} 5 ( ) 5 ( 1) i+1 0 i i X J = = 15 ( ) = (*) Let f : X Y be a bijection Prove that X is countably infinite if and only if Y is countably infinite For the first direction, suppose that Y is countably infinite Let g : Y N be a bijection Then g f : X N is a composition of bijections, so g is also a bijection, and hence X is countably infinite For the opposite direction, repeat the above argument using f 1 : Y X in place of f 4

5 1 Show directly (without using Proposition 4310) that N N is countably infinite (If you cannot derive a bijection, you may look to Problem 438 for one) Define f : N N N by f(a, b) = a (b + 1) 1 We will show that f is bijective For injectivity, suppose that f(a, b) = f(c, d) Then a (b+1) 1 = c (d + 1) 1, so a (b + 1) = c (d + 1) Define this value as k, and consider the unique prime decomposition of k Note that b + 1 and d + 1 are both odd, so the number of times that appears in the unique prime decomposition of k must be c, and also must be a Hence, a = c But if a = c, clearly b + 1 = d + 1, and hence b = d Therefore, (a, b) = (c, d), and f is injective For surjectivity, let k N If k is even, write k = m Then f(0, = 0 (m+1) 1 = m = k, and hence k is in the image of f If k is odd, then k + 1 is even Let a be the number of times that appears in the unique prime decomposition of k+1 Then k+1 is an odd integer, and hence there exists b such that a k+1 = b + 1 But then f(a, b) = a (b + 1) 1 = k = k, a and thus k is in the image of f Therefore, as f is both injective and surjective, it is bijective, so N N is countably infinite 13 (*) Let X be a finite set Show that N X = {f : X N} is countably infinite We work by induction on X If X = 1, so X = {x}, then all the functions on X are of the form f : X N, where f(x) = k We note that there is a bijection, F : N X N, by F (f) = f(x) Hence, X = N Now, suppose the result holds when X = n 1 Let X = n Let x X, and let X = X\{x}, so write X = {x} X and X = n 1 Given a function f : X N, define a function g : {x} N and h : X N by g(x) = f(x) and h(x ) = f(x ) Clearly, every f defines a unique g and h, and every choice of g and h correspond to exactly one f Hence, if we let F : N X (N {x} ) (N X ) by F (f) = (g, h), F is bijective Moreover, by the inductive hypothesis and the base case, we have that both N {x} and N X are countably infinite As finite products of countable sets are countable, we therefore have that N X is countably infinite By induction, then, the result holds for all possible X 5